# Can you solve this? The concentration of the Zn^(2+) ions in blood serum is 1.2 ppm. Express this amount in terms of millimole/L and milliequivalent/L.

Jul 10, 2015

The concentrations are $1.9 \cdot {10}^{- 2} \text{mM}$ and $3.8 \cdot {10}^{- 2} \text{meq/L}$.

#### Explanation:

So, you know that the ppm concentration of zinc cations, $Z {n}^{2 +}$, in blood serum is equal to 1.2 ppm.

A 1 ppm concentration means that you have 1 part solute, in your case zinc cations, per 1 million parts of solvent.

To get the ppm concentration, simply multiply the ratio that exists between the mass of the solute and the mass of the solvent by 1 million, or ${10}^{6}$.

To make calculations easier, assume that you're dealing with a 1-L sample of blood serum. The density of blood serum is know to be equal to $\text{1.025 g/mL}$, which means that your liter of serum will have a mass of

1cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1.025 g"/(1cancel("mL")) = "1.025 g"

Use the definition of ppm concentration to determine what mass of zinc cations you have in solution

$\text{ppm" = m_"solute"/m_"solvent} \cdot {10}^{6}$

m_"solute" = ("ppm" * m_"solvent")/10^6

${m}_{\text{solute" = (1.2 * "1025 g")/10^6 = 1.23 * 10^(-3)"g}}$

To express the concentration in milimoles per liter, you need to determine how many moles of $Z {n}^{2 +}$ you have in the sample. To do that, use its molar mass

1.23 * 10^(-3)cancel("g") * "1 mole"/(65.39cancel("g")) = 1.9 * 10^(-5)"moles" $Z {n}^{2 +}$

Expressed in milimoles, this is equal to

1.9 * 10^(-5)cancel("moles") * "1000 mmoles"/(1cancel("mole")) = 1.9 * 10^(-2)"mmoles" $Z {n}^{2 +}$

The concentration will thus be

$C = \frac{n}{V} = \left(1.9 \cdot {10}^{- 2} \text{mmoles")/("1 L") = color(green)(1.9 * 10^(-2)"mM}\right)$

To determine the concentration in miliequivalent per liter, you need to take into consideration the fact that normality only makes sense in the context of a chemical reaction.

An equivalent is simply a reactive unit - in your case, a reactive unit is a monovalent cation.

Since the zinc cation is actually a divalent cation, the number of equivalents will be equal to 2.

This means that you have

$N = \text{eq} \cdot C$, where

$\text{eq}$ - the number of equivalents;
$C$ - the molarity of the solution.

Plug in your values to get

N = 2 * 1.9 * 10^(-2)"mM" = color(green)(3.8 * 10^(-2)"meq/L")

SIDE NOTE I think that you can actually use the ppm concentration per liter of solution, which would make the mass of the zinc cations equal to

${m}_{\text{solute" = (1.2 * 1000)/10^6 = 1.2 * 10^(-3)"g}}$

As you can see, the values are not that different.